Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 35 / 3 / b / Solution

Let $q_0=n_0$, $q_1=n_0/c^2$ and $y=\overline{y}$. These are the two prior precision parameters. Completing the square in $n(\beta -y{)}^2+q_i{\beta }^2$ gives the Normal-normal conjugacy updateThus the posterior mean is a precision-weighted average of the observed mean and the prior center zero. The posterior variance is the reciprocal of the total precision.